Probability Mass Functions For Discrete Severity

Why Discrete Severity Needs Probability Masses

A loss severity model describes the amount of a loss, denoted by $X$. In some models, $X$ can take only particular values. For example, a hypothetical benefit might be paid in fixed increments, or a loss record might store amounts rounded to specified levels. In such a model, the probability at one possible loss amount can be positive. The appropriate tool is a probability mass function, or pmf.

The pmf answers a direct question: what is the probability that the loss is exactly $x$? Interval probabilities are then obtained by adding the masses for the possible values in the interval. The support and the interval endpoints both matter: an interval can contain several possible losses, or none at all.

Definition: A random variable $X$ is discrete if its possible values form a finite or countable set. Its probability mass function is $$ p_X(x)=P(X=x). $$ The support is the set of values to which the model assigns positive probability. A valid pmf satisfies $p_X(x)\geq 0$ for every $x$, and the probabilities over all possible values sum to 1: $$ \sum_{x\text{ in the support}} p_X(x)=1. $$

For any value outside the support, $P(X=x)=0$. A table is often the clearest way to specify a finite discrete severity model: one column lists possible loss amounts, and another lists the probability assigned to each amount. The probabilities must describe the entire distribution, not just a selected range of losses.

Checking and Using a Probability Mass Function

To check a proposed finite pmf, first verify that each entry is nonnegative, then add all entries. If the total is not 1, the table cannot be the complete pmf as written. It may contain an unknown normalizing constant, or it may omit probability assigned to another possible value. Do not calculate event probabilities from an invalid table as if it were already normalized.

For a point event, read its mass directly from the pmf. For an interval event, include exactly the support values that satisfy the stated inequalities. For example, $P(a\leq X<b)$ includes a support value equal to $a$, but excludes a value equal to $b$. The endpoints do not contribute probability merely because they are written in the interval; they contribute only if the inequalities include them and they are possible values.

Result: If $I$ is any interval and $X$ is discrete, then $$ P(X\in I)=\sum_{\substack{x\text{ in the support}\\x\in I}}p_X(x). $$ In particular, if the support values in $[a,b]$ are $x_1,\ldots,x_m$, then $P(a\leq X\leq b)=p_X(x_1)+\cdots+p_X(x_m)$.

This rule follows because the point events $\{X=x\}$ for distinct support values are mutually exclusive. The event $\{X\in I\}$ is the union of the point events for the support values inside $I$. Since disjoint event probabilities add, the interval probability is the sum of those masses. The same reasoning gives a useful connection with the cumulative distribution function $F_X(x)=P(X\leq x)$: the mass at $x$ is the increase in the cumulative probability at that point.

$$ p_X(x)=P(X=x)=F_X(x)-P(X<x). $$

For a discrete support arranged in increasing order, this means that consecutive cumulative probabilities can be subtracted to recover individual masses. This is especially useful when information is given as cumulative totals rather than as a pmf table.

Worked Example: Checking a Finite Loss Table

A hypothetical coverage model assigns the settled loss $X$, in dollars, one of four possible values: $0$, $400$, $900$, or $1{,}600$. The proposed probabilities are $0.12$, $0.28$, $0.37$, and $0.23$, respectively. Verify the pmf, then find $P(X=900)$, $P(400\leq X<1{,}600)$, and $P(X>900)$.

Method. Check nonnegativity and normalization first, then select the support values satisfying each event’s inequalities.

All four probabilities are nonnegative. Their sum is

$$ 0.12+0.28+0.37+0.23=1.00. $$

Thus the proposed table is a valid pmf on the stated support.

The point probability is the mass at $900$:

$$ P(X=900)=0.37. $$

For $400\leq X<1{,}600$, the included support values are $400$ and $900$. The value $0$ is below the interval, while $1{,}600$ is excluded by the strict upper inequality. Therefore,

P(400\leq X<1{,}600)=0.28+0.37=0.65.

For $X>900$, only $1{,}600$ qualifies, so

$$ P(X>900)=0.23. $$

Check. The event $400\leq X<1{,}600$ and the event $X>900$ are disjoint. Their probabilities sum to $0.65+0.23=0.88$, which is the probability of $X\geq400$; the omitted mass is $P(X=0)=0.12$. The endpoint handling is consistent.

Common mistake avoided: Including the $1{,}600$ mass in the first interval would change the upper condition from $X<1{,}600$ to $X\leq1{,}600$.

Normalizing Probabilities Given in Proportion

A model may specify relative weights rather than probabilities. In that case, each weight must be divided by the sum of all weights. Equivalently, if the weights are $w_1,\ldots,w_n$ and each probability is $cw_j$, the requirement that the probabilities sum to 1 determines $c$:

c\sum_{j=1}^{n}w_j=1 \qquad\Longrightarrow\qquad c=\frac{1}{\sum_{j=1}^{n}w_j}.

The resulting values are probabilities only after this normalization. An event probability is calculated from the normalized masses, not directly from the unscaled weights.

Worked Example: Normalizing Relative Severity Weights

For a hypothetical equipment-loss model, $X$ can be $0$, $250$, $750$, or $1{,}750$ dollars. The respective probabilities are proportional to $1:2:4:3$, so they are $c,2c,4c,3c$, where $c$ is an unknown constant. Find the pmf and calculate $P(250<X\leq1{,}750)$ and $P(X\geq750)$.

Method. Sum the relative weights to find the normalizing constant, verify the resulting masses, and then add the masses selected by each event.

The weights sum to $1+2+4+3=10$. The total probability must equal 1, so $10c=1$, giving $c=0.1$. The four probabilities are therefore

p_X(0)=0.10,\quad p_X(250)=0.20,\quad p_X(750)=0.40,\quad p_X(1{,}750)=0.30.

They are all nonnegative and their sum is $0.10+0.20+0.40+0.30=1.00$, as required.

The condition $250<X\leq1{,}750$ excludes $250$ because the lower inequality is strict, but includes $750$ and $1{,}750$. Thus

P(250<X\leq1{,}750)=0.40+0.30=0.70.

The event $X\geq750$ includes those same two support values, so

P(X\geq750)=0.40+0.30=0.70.

Check. The probability of the complementary event $X<750$ is $0.10+0.20=0.30$. Therefore $P(X\geq750)+P(X<750)=0.70+0.30=1.00$.

Common mistake avoided: Treating the weights $1,2,4,3$ as probabilities gives a total of 10, not 1. The normalization step is essential before interpreting any weight as a probability.

Recovering Point Masses From Cumulative Information

Sometimes a problem gives cumulative probabilities rather than point probabilities. Because $F_X(t)=P(X\leq t)$, the difference $F_X(t)-F_X(s)$ gives the probability of $s<X\leq t$. To recover a point mass at a support value, compare the cumulative probability at that value with the cumulative probability just before it. In a finite support table, this amounts to taking successive differences.

List the support in increasing order.
Do not assume every number between the smallest and largest values is possible.

Subtract consecutive cumulative totals.
The increase at a support value is the probability mass at that value.

Check the recovered masses.
They must be nonnegative and add to 1.

Apply the event’s endpoints.
Add only masses whose support values satisfy the event exactly.

Worked Example: Recovering Masses and an Interval Probability

A hypothetical loss model has support $100$, $300$, $700$, $1{,}200$, and $2{,}000$ dollars. It is specified that $P(X\leq100)=0.08$, $P(X\leq300)=0.25$, $P(X\leq700)=0.61$, $P(X\leq1{,}200)=0.88$, and $P(X\leq2{,}000)=1$. Recover the pmf and find $P(300<X\leq1{,}200)$ and $P(X\geq700)$.

Method. Take differences of consecutive cumulative totals to recover the mass at each support value.

There is no support value below $100$, so $p_X(100)=0.08$. At each later value, subtract the cumulative total at the preceding support value:

\begin{aligned} p_X(300)&=0.25-0.08=0.17,\\ p_X(700)&=0.61-0.25=0.36,\\ p_X(1{,}200)&=0.88-0.61=0.27,\\ p_X(2{,}000)&=1.00-0.88=0.12. \end{aligned}

The recovered masses are $0.08,0.17,0.36,0.27,0.12$. Each is nonnegative, and their sum is $0.08+0.17+0.36+0.27+0.12=1.00$.

The event $300<X\leq1{,}200$ includes $700$ and $1{,}200$, but not $300$. Hence

P(300<X\leq1{,}200)=0.36+0.27=0.63.

The event $X\geq700$ includes $700$, $1{,}200$, and $2{,}000$, giving

P(X\geq700)=0.36+0.27+0.12=0.75.

Check. Since $X$ has no support values strictly between $300$ and $700$, $P(300<X\leq1{,}200)$ also equals $F_X(1{,}200)-F_X(300)=0.88-0.25=0.63$.

Common mistake avoided: Using $F_X(1{,}200)$ alone would include the masses at $100$ and $300$. Subtracting $F_X(300)$ removes precisely those values excluded by the strict lower endpoint.

Common Traps and Practical Checks

A discrete pmf assigns probability to exact values. It is not a density. In a continuous model, an individual point has probability zero and interval probabilities are obtained by integrating a density. For a discrete model, the mass at a point can be positive and interval probabilities come from sums. Multiplying a pmf value by an interval width is not a valid way to calculate a discrete probability.

A frequent endpoint error is to include a boundary value in an interval even when its inequality is strict. Translate the event into a list of support values before adding; this simple step also prevents including values that are not possible under the model. Finally, use the normalization check as a safeguard: a complete pmf must total 1, including any mass at zero if zero is in the support.

Key takeaway: For discrete severity, the pmf gives the probability at each possible loss. Verify nonnegative masses summing to 1, then calculate any point or interval probability by selecting and adding exactly the relevant support masses.

Check Your Understanding

For each model, identify the support values relevant to the event before adding probabilities.

A hypothetical loss $X$ takes values $0$, $400$, and $900$ with probabilities $0.20$, $0.50$, and $0.30$. Find $P(X=400)$ and $P(0<X\leq900)$. (Answer: $0.50$ and $0.80$)
A loss takes values $100$, $300$, and $800$ with probabilities $0.10$, $0.45$, and $0.45$. Find $P(300\leq X<800)$ and $P(X>100)$. (Answer: $0.45$ and $0.90$)
For a model with support $0,250,500,1{,}000$, probabilities are proportional to $2:1:3:4$, in that order. Find the normalized probability $P(X\geq500)$. (Answer: $0.70$)
A model has support $0,1,2,3$, with respective probabilities $0.12,0.28,0.34,0.26$. Find $P(1<X\leq3)$ and $P(X=0\text{ or }X=3)$. (Answer: $0.60$ and $0.38$)
A loss has support $100,200,500,1{,}000,1{,}500$, with $P(X=100)=0.07$, $P(X\leq200)=0.18$, $P(X\leq500)=0.63$, and $P(X\leq1{,}000)=0.91$. The total probability is 1. Find $P(200<X\leq1{,}000)$. (Answer: $0.73$)

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